What are the solutions to the equation solvefor y,x^2+4xy+y^2=0 ?

The solutions to the equation solvefor y,x^2+4xy+y^2=0 are y=(sqrt(3)-2)x,y=-(2+sqrt(3))x

Find the zeros of solvefor y,x^2+4xy+y^2=0

The zeros of solvefor y,x^2+4xy+y^2=0 are y=(sqrt(3)-2)x,y=-(2+sqrt(3))x

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solve for

y, x^{2} + 4 x y + y^{2} = 0

y = (3 - 2) x, y = - (2 + 3) x

Solution steps

x^{2} + 4 x y + y^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} + 4 x y + x^{2} = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 x \pm ( 4 x ) ^{2} - 4 \cdot 1 \cdot x ^{2}}{2 \cdot 1}

Simplify

(4 x)^{2} - 4 \cdot 1 \cdot x^{2} : 23 x

y_{1, 2} = \frac{- 4 x \pm 2 3 x}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 x + 2 3 x}{2 \cdot 1}, y_{2} = \frac{- 4 x - 2 3 x}{2 \cdot 1}

y = \frac{- 4 x + 2 3 x}{2 \cdot 1} : (3 - 2) x

y = \frac{- 4 x - 2 3 x}{2 \cdot 1} : - (2 + 3) x

The solutions to the quadratic equation are:

y = (3 - 2) x, y = - (2 + 3) x

y, 2 x^{2} = y^{2}

4 x^{2} + 1 = 8 x - 2

x, x^{2} + 2 y^{2} + 3 z^{2} = 1

x^{2} = - 44

16.1 x^{2} + 4.398 x - 2.661 = 0

What are the solutions to the equation solvefor y,x^2+4xy+y^2=0 ?
The solutions to the equation solvefor y,x^2+4xy+y^2=0 are y=(sqrt(3)-2)x,y=-(2+sqrt(3))x
Find the zeros of solvefor y,x^2+4xy+y^2=0
The zeros of solvefor y,x^2+4xy+y^2=0 are y=(sqrt(3)-2)x,y=-(2+sqrt(3))x